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L1 = L3 = 0.356 A AP = 1.09 B F L2 = 0.785 0.544 02 96° L4 = 0.950 04 1

L1 = L3 = 0.356 A AP = 1.09 B F L2 = 0.785 0.544 02 96° L4 = 0.950 04 1

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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The fourbar linkage shown is made of Aluminum whose density is 2,170 kg/m3.
Each link is 5 cm wide and 1 cm thick. In the position shown, the crank has an
angular velocity of 15 rad/s and an angular acceleration of -10 rad/s2. A vertical
force acting downward at point P is 100 N. Link lengths are in meters. The center
of gravity of each link lies at the center of the link along its length.

You may ignore rounded link ends when calculating the moment of inertia.
Find the joint forces and the torque required to drive the mechanism at this instant using Newtonian method. 

Hint: T12 ≈ 12.73 Nm and F43 ≈ [−19.83    113.81]N

L1 = L3 = 0.356 A AP = 1.09 B F L2 = 0.785 0.544 02 96° L4 = 0.950 04 1
Transcribed Image Text:L1 = L3 = 0.356 A AP = 1.09 B F L2 = 0.785 0.544 02 96° L4 = 0.950 04 1
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